MathsCBSE Class 8 · Algebra Play

Largest Products and Divisibility Tricks Decoded for Class 8

Updated 17 October 2026 · Fact-checked

To make the largest product of the form (two-digit) x (one-digit) from three digits, use the largest digit as the multiplier and the other two in decreasing order. Algebra explains digit tricks: ab - ba = 9(a - b), ab + ba = 11(a + b), abc + bca + cab = 111(a + b + c) = 37 x 3 (a + b + c), and abcabc = abc x 1001 = abc x 7 x 11 x 13.

In each pair the larger multiplicand wins, and the overall winner 32 × 5 = 160 uses the largest digit 5 as the multiplier.

Illustration drawn for QuestionBuddy.

Understand Largest Products and Divisibility Tricks Decoded

Largest product. Fill the digits 2, 3, 5 once each into $□□ \times □$ . The six products are $23 \times 5$ , $25 \times 3$ , $32 \times 5$ , $35 \times 2$ , $52 \times 3$ and $53 \times 2$ . Group them by multiplier: in each pair the larger multiplicand wins, leaving $53 \times 2$ , $52 \times 3$ and $32 \times 5$ . Then $52 \times 3 = 15 \times 10 + 2 \times 3$ and $32 \times 5 = 15 \times 10 + 2 \times 5$ have the same first term and $32 \times 5$ has the larger second term, so the largest product is $32 \times 5 = 160$ .

*General proof.* For digits $p < q < r$ : group by the multiplier; with the larger tens digit in each pair, three products remain: $r q \times p$ , $r p \times q$ and $q p \times r$ . Since $q > p$ , $r p \times q > r q \times p$ : indeed $(10 r + p) q = 10 r q + pq$ and $(10 r + q) p = 10 r p + pq$ have the same second term, and $r q > r p$ because $q > p$ . Between $q p \times r = 10 q r + p r$ and $r p \times q = 10 r q + pq$ , the first terms are equal and $p r > pq$ because $r > q$ . So the largest digit is the multiplier and the other two digits, in decreasing order, form the multiplicand: $q p \times r$ .

Digit tricks with algebra. Write a two-digit number with tens digit $a$ and units digit $b$ as $10 a + b$ (written $ab$ ).

*Reversal difference:* $ab - ba = (10 a + b) - (10 b + a) = 9 a - 9 b = 9 (a - b)$ , so the difference of a two-digit number and its reverse is always a multiple of 9 (and the quotient is $a - b$ ). If $a < b$ , use $ba - ab = 9 (b - a)$ instead.
*Reversal sum:* $ab + ba = 11 a + 11 b = 11 (a + b)$ , always a multiple of 11. For example, $28 + 82 = 110 = 11 \times 10$ .
*Cyclic sum:* for a three-digit number $ab c = 100 a + 10 b + c$ , the numbers $b c a$ and $c ab$ are made by cycling the digits. Their sum is $111 a + 111 b + 111 c = 111 (a + b + c) = 37 \times 3 (a + b + c)$ . So the sum is divisible by 37 and also by 3 (and by 111). Example: $247 + 472 + 724 = 1443 = 111 \times 13 = 37 \times 39$ .
*Repeated digits:* a six-digit number $ab c ab c = 1000 \times ab c + ab c = 1001 \times ab c$ , and $1001 = 7 \times 11 \times 13$ . So dividing $ab c ab c$ by 7, then 11, then 13 gives back $ab c$ . Example: $465465 \div 7 \div 11 \div 13 = 465$ .

Word problems with equations. The chapter also uses algebra for ordinary puzzles, such as heads and legs of horses and hens (suppose all 55 animals are hens and compare the legs), ages in years (a mother 5 times her daughter's age now and 3 times in 6 years), and doubling flowers at shrines. Always name the unknowns, translate each statement into an equation and solve.

Number tricks, pyramids and grids are on the previous topic page of this chapter.

Largest Products and Divisibility Tricks Decoded: formulas to remember

Largest product rule: $p < q < r \Rightarrow max = (10 q + p) \times r$; Largest digit multiplies the number with the other digits in decreasing order.
Reversal difference: $ab - ba = 9 (a - b)$; Multiple of 9.
Reversal sum: $ab + ba = 11 (a + b)$; Multiple of 11.
Cyclic sum: $ab c + b c a + c ab = 111 (a + b + c)$; 111 = 3 x 37.
Repeated three digits: $ab c ab c = 1001 \times ab c, 1001 = 7 \times 11 \times 13$; So divisible by 7, 11 and 13.

How to solve Largest Products and Divisibility Tricks Decoded questions

Turn digit statements into place-value expressions.

1Write each number as a sum of place values: ab = 10a + b, abc = 100a + 10b + c.
2Form the sum, difference or product demanded by the trick and expand.
3Collect like terms and factorise to expose the divisor (9, 11, 37, 1001).
4For maximum or minimum problems, group the options by the common factor, compare two at a time and use expansion for close cases.
5Check with one numerical example.
6State the conclusion in words (always divisible by 9, and so on).

Largest Products and Divisibility Tricks Decoded in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Largest product

Suppose a quiz asks you to use the digits 2, 6 and 8 once each to make the largest possible product of a two-digit number and a one-digit number.

What is the largest product and how do you know?

Try it first, then show the solutionSolution

Largest digit as multiplier: 8. The other digits 6 and 2 in decreasing order: 62. Product $62 \times 8 = 496$ .
The closest rival is $82 \times 6 = 492$ . Compare: $62 \times 8 = 480 + 16$ and $82 \times 6 = 480 + 12$ .

Answer: 496

Memory hook: Biggest digit multiplies; the rest go down in order.

A divisibility trick

Imagine a friend who thinks of a three-digit number, writes it twice side by side to make a six-digit number and tells you it is 465465.

What was the original number, and how can you find it by dividing only by 7, 11 and 13?

Try it first, then show the solutionSolution

A six-digit number $ab c ab c = 1001 \times ab c$ and $1001 = 7 \times 11 \times 13$ .
$465465 \div 7 = 66495$ , $66495 \div 11 = 6045$ , $6045 \div 13 = 465$ . The original number is 465.

Answer: 465

Memory hook: 1001 is 7 x 11 x 13: dividing by all three undoes the repetition.

Tricks to remember Largest Products and Divisibility Tricks Decoded

Short memory aids. Say each one aloud once, then use it on a question the same day.

Biggest digit multiplies
For two-digit times one-digit, give the biggest digit to the multiplier and arrange the other two in decreasing order.
Use it when: Largest-product questions.
Reverse and subtract: nine; reverse and add: eleven
ab - ba is a multiple of 9; ab + ba is a multiple of 11.
Use it when: Two-digit digit tricks.
111 = 3 x 37
Cycling the digits of a three-digit number and adding gives 111 times the digit sum.
Use it when: Divisibility by 37 and 3.
1001 = 7 x 11 x 13
Repeating a three-digit number multiplies it by 1001.
Use it when: The abcabc trick.

Largest Products and Divisibility Tricks Decoded: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Largest product · 1 mark · MCQBoard level

Using the digits 1, 3 and 7 once each in a box pattern (two-digit number) x (one-digit number), the largest product is: (a) 217 (b) 213 (c) 91 (d) 73

Try it first, then show the solutionSolution

Largest digit 7 is the multiplier; the other digits in decreasing order give 31. Product $31 \times 7 = 217$ .
The next best, $71 \times 3 = 213$ , is smaller.

Answer: (a) 217

Example 2: Reversal difference · 2 marks · Short answerBoard level

Find 83 - 38 and show with algebra that the difference of any two-digit number and its reverse (with tens digit larger) is 9 times the difference of its digits.

Try it first, then show the solutionSolution

$83 - 38 = 45 = 9 \times 5$ , and the digits differ by $8 - 3 = 5$ . (1 mark)
In general $(10 a + b) - (10 b + a) = 9 a - 9 b = 9 (a - b)$ . (1 mark)

Answer: 45 = 9 x 5; in general 9(a - b)

Example 3: Largest product with justification · 3 marks · Short answerHarder board level

Find the largest product that can be made by placing the digits 2, 6 and 8 in a two-digit x one-digit pattern, and justify why 82 x 6 is smaller.

Try it first, then show the solutionSolution

Multiplier 8, the other digits in decreasing order: $62 \times 8 = 496$ . (1 mark)
The rival with multiplier 6 and the larger multiplicand is $82 \times 6$ . (1 mark)
$62 \times 8 = 60 \times 8 + 2 \times 8 = 480 + 16$ and $82 \times 6 = 80 \times 6 + 2 \times 6 = 480 + 12$ ; the first terms are equal and $16 > 12$ . (1 mark)

Answer: 496; the rival 82 x 6 = 492 is smaller

Example 4: Reversal sum · 3 marks · Short answerHarder board level

Show that the sum of a two-digit number and its reverse is a multiple of 11. Find all two-digit numbers with different non-zero digits whose sum with their reverse is 99.

Try it first, then show the solutionSolution

$(10 a + b) + (10 b + a) = 11 a + 11 b = 11 (a + b)$ . (1 mark)
For 99: $11 (a + b) = 99$ , so $a + b = 9$ . (1 mark)
With non-zero digits: 18, 27, 36, 45, 54, 63, 72, 81. (1 mark)

Answer: 11(a + b); the numbers are 18, 27, 36, 45, 54, 63, 72, 81

Example 5: Cyclic sum · 4 marks · Long answerHarder board level

For a three-digit number abc, show that abc + bca + cab is divisible by 37 and by 3. Use 247 as an example.

Try it first, then show the solutionSolution

$ab c + b c a + c ab = (100 a + 10 b + c) + (100 b + 10 c + a) + (100 c + 10 a + b)$ . (1 mark)
$= 111 a + 111 b + 111 c = 111 (a + b + c)$ . (1 mark)
$111 = 3 \times 37$ , so the sum equals $37 \times 3 (a + b + c)$ , divisible by 37 and by 3. (1 mark)
For 247: $247 + 472 + 724 = 1443 = 111 \times 13 = 37 \times 39$ . (1 mark)

Answer: 111(a + b + c) = 3 x 37 x (a + b + c); for 247 the sum is 1443 = 37 x 39

Where students lose marks in Largest Products and Divisibility Tricks Decoded

The slips examiners see most often, and what to write instead.

Mistake: Writing the two-digit number ab as a x b
ab means 10a + b, not a times b.
Wrong working
"ab - ba = ab - ba = 0."
Right working
$(10 a + b) - (10 b + a) = 9 (a - b)$ .
Do this: Expand with place values first.
Usually costs: The whole proof.
Mistake: Forgetting that 111 also gives divisibility by 3
111 = 3 x 37; the sum is 3 x 37 x (a + b + c).
Wrong working
"The sum is divisible by 37 only."
Right working
$111 (a + b + c) = 3 \times 37 \times (a + b + c)$ , divisible by both 3 and 37.
Do this: Factorise 111 before concluding.
Usually costs: 1 mark.
Mistake: Choosing the largest two digits for the multiplicand
The largest digit should multiply; the two smaller form the multiplicand.
Wrong working
"Digits 2, 6, 8: 86 x 2 = 172 is the largest."
Right working
$62 \times 8 = 496$ is larger.
Do this: Test two or three arrangements numerically.
Usually costs: The answer.
Mistake: Not verifying with an example
An example catches algebra slips.
Wrong working
"ab + ba = 10(a + b)"
Right working
$ab + ba = 11 (a + b)$ ; check $28 + 82 = 110 = 11 \times 10$ .
Do this: Substitute small digits after expanding.
Usually costs: 1 mark.

Exam tips

Expand every digit-trick with place values (100a + 10b + c) as the first line.
Factorise the final expression so the divisor appears clearly.
Support the algebra with a numerical example in the last line.
For largest-product problems, show the grouping argument and one close comparison by expansion.
Write the conclusion in words: always divisible by 9, 11, 37 or 1001.

Practice questions on Largest Products and Divisibility Tricks Decoded

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Assertion-reasonBoard level
Assertion (A): Any six-digit number of the form abcabc is divisible by 7, 11 and 13.
Reason (R): abcabc = abc x 1001 and 1001 = 7 x 11 x 13.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Check 7 x 11 x 13.
Answer
(a). $7 \times 11 = 77$ and $77 \times 13 = 1001$ , so $ab c ab c = ab c \times 7 \times 11 \times 13$ .
Q2 · 2 marks · Very short answerBoard level
Divide 465465 by 7, then by 11, then by 13. What do you get, and why?
Hint
465465 = 465 x 1001.
Answer
$465465 \div 7 = 66495$ , $\div 11 = 6045$ , $\div 13 = 465$ . This works because $465465 = 465 \times 1001$ and $1001 = 7 \times 11 \times 13$ .
Q3 · 3 marks · Short answerHarder board level
Find the largest product obtained by placing the digits 4, 7 and 9 in a two-digit x one-digit pattern. Compare with the closest rival using expansion.
Hint
Multiplier 9, then 7 and 4 in decreasing order.
Answer
$74 \times 9 = 666$ . The closest rival is $94 \times 7 = 658$ . By expansion: $74 \times 9 = 70 \times 9 + 4 \times 9 = 630 + 36$ and $94 \times 7 = 90 \times 7 + 4 \times 7 = 630 + 28$ ; the first terms are equal and $36 > 28$ .
Q4 · 4 marks · Long answerHOTS / competency
For digits p < q < r, prove that (10q + p) x r is greater than (10r + p) x q and greater than (10r + q) x p. Check with p = 2, q = 6, r = 8.
Hint
Expand each product and compare.
Answer
$(10 q + p) r = 10 q r + p r$ . Against $(10 r + p) q = 10 q r + pq$ : the first terms are equal and $p r > pq$ since $r > q$ . So the first is greater. Against $(10 r + q) p = 10 p r + pq$ : the difference is $(10 q r + p r) - (10 p r + pq) = 10 r (q - p) + p (r - q)$ , which is positive since $q > p$ and $r > q$ . Check: $(10 \cdot 6 + 2) \cdot 8 = 496$ ; $(10 \cdot 8 + 2) \cdot 6 = 492$ ; $(10 \cdot 8 + 6) \cdot 2 = 172$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case study: a magician asks a student to write a two-digit number with different non-zero digits, reverse the digits, subtract the smaller number from the larger, and tell the result.
(i) Show that the result is always a multiple of 9. (1 mark)
(ii) The result is 54. List all the numbers the student could have started with. (2 marks)
(iii) Could the result be 40? Explain. (1 mark)
Hint
The result is 9(a - b); 54 = 9 x 6.
Answer
(i) $∣ ab - ba ∣ = 9∣ a - b ∣$ , a multiple of 9. (ii) $9∣ a - b ∣ = 54$ gives $∣ a - b ∣ = 6$ with non-zero digits: (7, 1), (8, 2), (9, 3): the numbers are 71, 17, 82, 28, 93, 39. (iii) No: 40 is not a multiple of 9.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is ab - ba (a > b)?Show answer
9(a - b).
2.What is ab + ba?Show answer
11(a + b).
3.What is abc + bca + cab?Show answer
111(a + b + c) = 3 x 37 x (a + b + c).
4.What is abcabc?Show answer
abc x 1001, and 1001 = 7 x 11 x 13.
5.Largest product with three digits in a two-digit x one-digit pattern?Show answer
Largest digit as multiplier, the other two in decreasing order.
6.Is ab the same as a x b?Show answer
No, ab = 10a + b.

Largest Products and Divisibility Tricks Decoded: frequently asked questions

Why is the largest digit the multiplier?

Because the grouping argument shows that among the three finalists the first terms are equal, and the second term is largest when the biggest digit multiplies the unit digit of the multiplicand.

Does the reversal trick work if the units digit is 0?

Algebraically yes, but the reverse is then not a two-digit number (for example 50 reversed is 05), so questions usually require non-zero digits.

Is a number made by cycling digits always divisible by 37?

The sum of the three cyclic numbers is; each cyclic number on its own need not be.

Why do these tricks work for every number?

Because the algebra uses letter-numbers a, b, c for any digits, so the proof holds for all digits.